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A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 kg directly onto the middle of the wheel. The clay is in the shape of a pancake and has a radius of 0.19 m. Assume both the wheel and the chunk of clay can be modeled as solid cylinders (I = ½ MR2 ). What is the new tangential velocity of the wheel and the clay?

Respuesta :

xero099

Answer:

[tex]v_{f,w} = 1.791\,\frac{m}{s}[/tex], [tex]v_{f,c} = 0.972\,\frac{m}{s}[/tex]

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

[tex]I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}[/tex]

The final angular speed is:

[tex]\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}[/tex]

[tex]\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)[/tex]

[tex]\omega_{f} \approx 5.116\,\frac{rad}{s}[/tex]

The tangential velocities of the wheel and the clay are, respectively:

[tex]v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )[/tex]

[tex]v_{f,w} = 1.791\,\frac{m}{s}[/tex]

[tex]v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )[/tex]

[tex]v_{f,c} = 0.972\,\frac{m}{s}[/tex]

oladayoademosu

The tangential velocity of the wheel is 1.79 m/s and the tangential velocity of the clay is 0.97 m/s

From the law of conservation of angular momentum, initial angular momentum = final angular momentum

L = L' + L'' where L = initial angular momentum of wheel, L' = final angular momentum of wheel, and L'' = angular momentum of clay.

Iω = Iω' + I'ω' where

  • I = rotational inertia of wheel = 1/2MR² where
  • M = mass of wheel = 4.2 kg and R = radius of wheel = 0.35 m,
  • ω = initial angular speed of wheel = 0.98 rev/s = 0.98 × 2π rad/s = 6.16 rad/s,
  • I' = rotational inertia of clay = 1/2mr² where
  • m = mass of clay = 2.9 kg and r = radius of clay = 0.19 m,
  • ω = final angular speed of wheel + clay

Angular speed of wheel + clay

So, making ω' subject of the formula, we have

ω'  = Iω/(I + I')

ω'  = Iω/(I + I')

ω'  = 1/2MR²ω/(1/2MR² + 1/2mr²)

ω'  = MR²ω/(MR² + mr²)

Substituting the values of the variables into the equation, we have

ω'  = 4.2 kg × (0.35 m)² × 6.16 rad/s/(4.2 kg × (0.35 m)²  + 2.9 kg × (0.19 m)² )

ω'  = 4.2 kg × 0.1225 m² × 6.16 rad/s/(4.2 kg × 0.1225 m²  + 2.9 kg × 0.0361 m² )

ω'  = 0.5145 kgm² × 6.16 rad/s/(0.5145 kg m²  + 0.1047 kgm)²

ω'  = 3.169kgm²rad/s ÷0.6192kgm²

ω' = 5.12 rad/s

The tangential velocity of the wheel

Since tangential speed v = Rω' where

  • R = radius of wheel = 0.35 m and
  • ω' = angular speed = 5.12 rad/s

So, substituting these into the equation, we have

v = Rω'

v = 0.35 m × 5.12 rad/s

v = 1.792 m/s

v ≅ 1.79 m/s

The tangential velocity of the wheel is 1.79 m/s

The tangential velocity of the clay

Since tangential velocity v' = rω' where

  • r = radius of clay = 0.19 m and
  • ω' = angular speed = 5.12 rad/s

So, substituting these into the equation, we have

v = rω'

v = 0.19 m × 5.12 rad/s

v = 0.973 m/s

v' ≅ 0.97 m/s

The tangential velocity of the clay is 0.97 m/s

The tangential velocity of the wheel is 1.79 m/s and the tangential velocity of the clay is 0.97 m/s

Learn more about tangential velocity here:

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